Approximate solution to a transcendental equation I'm working on a physics problem and stumbled upon the following equation:
$$h=\frac{2n\pi+\arctan\left(\frac{c}{b}\right)}{b}$$
where $n \in \mathbb{Z}$, $c \in [0,20]$ and $h \in \mathbb{R}^+$. This equation has to be solved for $b$, which can be done numerically ofcourse, but if possible, I would love to have a quite accurate expression in the given interval. Ive already did some work and obtained the following fair approximation for $b$:
$$b\approx\frac{(1+2n)\pi c}{2+ch}$$ However, this equation breaks down at low $c$. Thus, I was wondering if more accurate (analytical) expressions are possible.
 A: Let $\frac c b=x$ to make the equation
$$2 n\pi   x+x \tan ^{-1}(x)=k \qquad \text{with} \qquad k=ch$$ Expand the lhs as a Taylor series around $x=0$
$$2 n\pi   x+x \tan ^{-1}(x)=2 n\pi   x+x^2-\frac{x^4}{3}+\frac{x^6}{5}+O\left(x^8\right)$$ Use series reversion
$$x=\frac{k}{2 \pi  n}-\frac{k^2}{8 \pi ^3 n^3}+\frac{k^3}{16 \pi ^5
   n^5}+\frac{k^4 \left(4 \pi ^2 n^2-15\right)}{384 \pi ^7 n^7}+\frac{k^5 \left(7-4
   \pi ^2 n^2\right)}{256 \pi ^9 n^9}+O\left(k^6\right)$$
In view of this results, we can also perform one single iteration iteration of Newton method
$$x_0=\frac k {2 n\pi}\implies x_1=\frac k {2 n\pi}-\frac{k \cot ^{-1}\left(\frac{2 \pi  n}{k}\right)}{2 \pi  n \left(2 \pi  n
   \left(\frac{k}{k^2+4 \pi ^2 n^2}+1\right)+\cot ^{-1}\left(\frac{2 \pi 
   n}{k}\right)\right)}$$
Edit (after @River Li's answer)
We can expand the function
$$f(b)=b-\frac{\tan ^{-1}\left(\frac{c}{b}\right)+2 \pi  n}{h}$$ as series solution around $b=\frac {2n\pi}h$,  use series reversion and have, as an approximation
$$b=\frac {2n\pi}h+t+\sum_{k=2}^p a_k \,t^k+ O(t^{p+1})$$ where $$t=\frac{c^2 h^2+4 \pi ^2 n^2}{h \left(c^2 h^2+c h+4 \pi ^2 n^2\right)}\,\tan ^{-1}\left(\frac{c h}{2 \pi  n}\right)$$ The first coefficient is
$$a_2=\frac{2 \pi  c h^2 n}{\left(c^2 h^2+4 \pi ^2 n^2\right) \left(c^2 h^2+c h+4 \pi ^2   n^2\right)}$$
Repeating @River Li's test cases, the second term is perfectly negligible.
What we can also do is one iteration of Halley method using $b_0=\frac {2n\pi}h$ and have as estimate
$$b=\frac {2n\pi}h+\frac{\left(c^2 h^2+4 \pi ^2 n^2\right) \left(c h (c h+1)+4 \pi ^2 n^2\right) \cot
   ^{-1}\left(\frac{2 \pi  n}{c h}\right)}{h \left(\left(c h (c h+1)+4 \pi ^2
   n^2\right)^2-2 \pi  c h n \cot ^{-1}\left(\frac{2 \pi  n}{c h}\right)\right)}$$
A: We deal with the case $c > 0, h > 0, n\ge 1$.
The equation is written as
$$b = \frac{2n\pi}{h} + \frac{\arctan\frac{c}{b}}{h}.$$
Clearly, the equation has a unique real solution $b$. More precisely,
this solution is located in $ (\frac{2n\pi}{h}, \frac{2n\pi + \pi/2}{h})$.
Using the famous Lagrange inversion theorem, the solution is given by
$$b = \frac{2n\pi}{h} + \sum_{k=1}^\infty \frac{1}{k! h^k}
\frac{\partial^{k - 1}}{\partial x^{k - 1}}\left[\left(\arctan\frac{c}{x}\right)^k\right]\Bigg\vert_{x = \frac{2n\pi}{h} }.$$
An approximation is
\begin{align*}
 b &\approx \frac{2n\pi}{h} + \sum_{k=1}^2 \frac{1}{k! h^k}
 \frac{\partial^{k - 1}}{\partial x^{k - 1}}\left[\left(\arctan\frac{c}{x}\right)^k\right]\Bigg\vert_{x = \frac{2n\pi}{h} }\\
 &= \frac{2n\pi}{h} + \frac{1}{h}\arctan \frac{ch}{2n\pi}
 - \frac{c}{4\pi^2 n^2 + c^2h^2}\arctan \frac{ch}{2n\pi}.
\end{align*}
Examples:
(1) $c = 1/2, h = 1, n = 2$,
the solution is $\approx 12.60601344$,
and the approximation is $12.60601266$.
(2) $c = 1/100, h = 1/5, n = 10$,
the solution is $\approx 314.1594245$,
and the approximation is $314.1594246$.
(3) $c = 1/10, h = 100, n = 10$,
the solution is $\approx 0.6298929625$,
and the approximation is $0.6298929436$.
A: HINT.-Let $\arctan(\frac cb)=\theta$ so $\tan(\theta)=\dfrac cb$.
$$\tan(bh)=\frac{\tan(2n\pi)+\tan(\theta))}{1-\tan(2n\pi)\tan(\theta)}=\frac{\frac cb}{1}$$
$$\tan(bh)=hb+\dfrac{h^3b^3}{3}+\dfrac{2h^5b^5}{15}+O(b^6)$$ it follows the easier equation
$$\frac cb=hb+\dfrac{h^3b^3}{3}+\dfrac{2h^5b^5}{15}+O(b^6)$$
